Question

If \[\left| \begin{array}{l} x - 4\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,x - 4\,\, \end{array} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},\]         then the ordered pair $$(A, B)$$  is equal to:

A. $$(- 4, 3)$$
B. $$(- 4, 5)$$  
C. $$(4, 5)$$
D. $$(- 4, - 5)$$
Answer :   $$(- 4, 5)$$
Solution :
\[{\rm{Here, }}\left| \begin{array}{l} x - 4\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,x - 4 \end{array} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}\]
$${\text{Put }}x = 0$$
\[ \Rightarrow \,\,\left| \begin{array}{l} - 4\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,0\,\,\,\,\,\,\, - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - 4 \end{array} \right| = {A^3}\]
$$\eqalign{ & \Rightarrow \,\,{A^3} = {\left( { - 4} \right)^3} \cr & \Rightarrow \,\,A = - 4 \cr} $$
\[ \Rightarrow \,\,\left| \begin{array}{l} x - 4\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,x - 4 \end{array} \right| = \left( {Bx - 4} \right){\left( {x + 4} \right)^2}\]
Now take $$x$$ common from both the sides
\[\therefore \,\,\left| \begin{array}{l} 1 - \frac{4}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \frac{4}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \frac{4}{x} \end{array} \right| = \left( {B - \frac{4}{x}} \right){\left( {1 + \frac{4}{x}} \right)^2}\]
$${\text{Now take }}x \to \infty ,\,{\text{then }}\frac{1}{x} \to 0$$
\[ \Rightarrow \,\,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 \end{array} \right| = B\]
$$ \Rightarrow \,\,B = 5$$
∴ Ordered pair $$(A, B)$$  is $$(- 4, 5)$$

Releted MCQ Question on
Algebra >> Matrices and Determinants

Releted Question 1

Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$  be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$  be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then

A. $$C$$ is empty
B. $$B$$  has as many elements as $$C$$
C. $$A = B \cup C$$
D. $$B$$  has twice as many elements as elements as $$C$$
Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
Releted Question 3

Let $$a, b, c$$  be the real numbers. Then following system of equations in $$x, y$$  and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$    $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$     has

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$

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Matrices and Determinants


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